What volume of 6.58 M HNO3 is required to
prepare 671 mL of a 2.7 M HNO3 solution?
Answer in units of mL
To determine the volume of 6.58 M HNO3 required to prepare a 2.7 M HNO3 solution, we can use the concept of dilution.
The formula for dilution is:
C1V1 = C2V2
where C1 and V1 represent the concentration and volume of the initial solution, and C2 and V2 represent the concentration and volume of the final solution.
In this case, we have:
C1 = 6.58 M (concentration of the initial solution)
V1 = ? (volume of the initial solution)
C2 = 2.7 M (concentration of the final solution)
V2 = 671 mL (volume of the final solution)
Plugging in the values into the dilution formula, we get:
6.58 M * V1 = 2.7 M * 671 mL
Now we can solve for V1:
V1 = (2.7 M * 671 mL) / 6.58 M
V1 ≈ 275.30 mL
Therefore, approximately 275.30 mL of the 6.58 M HNO3 solution is required to prepare 671 mL of a 2.7 M HNO3 solution.
Use the dilution formula.
M1V1 = M2V2